Question

Calculate the equilibrium composition for the reaction H2−− 1 2O2 −−)−−* H2O when the ratio of the number of moles of elemental hydrogen to elemental oxygen is unity. Perform this calculation at T = 2000 K and P = 1 atm.

Answer 1

Answer:

the equilibrium composition Kp is 3471.01

Explanation:

Given;

H₂ + ½O₂ ↔ H₂O

Using Gibb's formula;

$lnK_p = -\frac{\delta G_{(T)}}{RT}$---------(i)

where;

ΔG(T) is change in Gibb's energy and it is given as;

$-\delta G_{(T)} = V_{H_2O}*g_{H_2O}(T)-V_{H_2}*g_{H_2}(T)-V_{O_2}*g_{O_2}(T)\\\\=V_{H_2O}(h-Ts)_{H_2O -V_{H_2}(h-Ts)_{H_2}-V_{O_2}(h-Ts)_{O_2}$

$= V_{H_20[h_f+h_{2000}-hs)-Ts]H_2O - V_{H_2}[h_f+h_{2000}-hs)-Ts]H_2- V_{O_2}[h_f+h_{2000}-hs)-Ts]O_2$

= 1(-241,820 + 82,593 -9904 - 2000 x 264.571) - 1(0 + 61,400 - 8468 - 2000 x 188.297) - 0.5(0 + 67,881 -8682 - 2000 x 268.655)

= -698273 + 323662 + 239055.5

= -135,555.5 KJ

Substituting this value into equation (i)

$lnK_p = -\frac{\delta G_{(T)}}{RT} = lnK_p = \frac{135,555.5}{8.314*2000}\\\\lnK_p = 8.1522\\\\K_p = e^{8.1522} = 3471.01$

Therefore, the equilibrium composition Kp is 3471.01